1504 CHAPTER 45. TRACES OF SOBOLEV SPACES

I will write U instead of U ∩B× (a,b) to save space but this does not matter because u isassumed to be zero outside the indicated region. Then∫

Rn

∫Rn

|Eu(x̂,xn)−Eu(ŷ,yn)|p∣∣∣|x̂− ŷ|2 +(xn− yn)2∣∣∣(1/2)(n+pθ)

dxdy

=∫

U

∫U

|u(x̂,xn)−u(ŷ,yn)|p∣∣∣|x̂− ŷ|2 +(xn− yn)2∣∣∣(1/2)(n+pθ)

dxdy+

∫U+

∫U

|u(x̂,xn)−Eu(ŷ,yn)|p∣∣∣|x̂− ŷ|2 +(xn− yn)2∣∣∣(1/2)(n+pθ)

dxdy+ (45.3.17)

∫U

∫U+

|Eu(x̂,xn)−u(ŷ,yn)|p∣∣∣|x̂− ŷ|2 +(xn− yn)2∣∣∣(1/2)(n+pθ)

dxdy+ (45.3.18)

∫U+

∫U+

|Eu(x̂,xn)−Eu(ŷ,yn)|p∣∣∣|x̂− ŷ|2 +(xn− yn)2∣∣∣(1/2)(n+pθ)

dxdy (45.3.19)

Consider the second of the integrals on the right of the equal sign. Using Fubini’s theorem,it equals ∫

U

∫U+

|u(x̂,xn)−u(ŷ,2g(ŷ)− yn)|p∣∣∣|x̂− ŷ|2 +(xn− yn)2∣∣∣(1/2)(n+pθ)

dydx

=∫

U

∫B

∫∞

g(ŷ)

|u(x̂,xn)−u(ŷ,2g(ŷ)− yn)|p∣∣∣|x̂− ŷ|2 +(xn− yn)2∣∣∣(1/2)(n+pθ)

dyndŷdx

=∫

U

∫B

∫ g(ŷ)

−∞

|u(x̂,xn)−u(ŷ,zn)|p∣∣∣|x̂− ŷ|2 +(xn− (2g(ŷ)− zn))2∣∣∣(1/2)(n+pθ)

dzndŷdx

I need to estimate |xn− zn| .

|xn− zn| ≤ |xn−g(x̂)|+ |g(x̂)−g(ŷ)|+ |g(ŷ)− zn|

≤ g(x̂)− xn +K |x̂− ŷ|+ yn−g(ŷ)≤ |yn− xn|+2K |x̂− ŷ|

and so

|xn− zn|2 ≤ 8K2 |x̂− ŷ|2 +2 |yn− xn|2

≤ |x̂− ŷ|2 +2 |yn− xn|2